Study skills

The Power of Checking

Mathematics is almost unique among school subjects in offering students a genuine ability to verify their own answers. Write a history essay and you cannot know whether it is correct — your teacher will decide. Solve a maths problem and, in most cases, you can find out for yourself before anyone marks it. That is an extraordinary advantage, and most students do not use it enough.

Why maths is different

The key insight is that the method you use to check an answer is usually completely different from the method you used to find it. This is what makes a check genuinely informative: two independent routes to the same conclusion give you strong evidence that the conclusion is correct. If you simply repeat the same calculation, you are likely to repeat the same mistake.

A good check takes a fraction of the time of the original solution, and the habit of checking — systematically, not just glancing over your working — is one of the most reliable ways to improve your exam performance.

Back-substitution

If you have solved an equation, put your answer back into the original equation and see whether it is satisfied. This works for algebraic equations, differential equations, and systems of simultaneous equations alike.

Example — a quadratic

Suppose you solve x² − 5x + 6 = 0 and obtain x = 2 and x = 3. Check:

Try it yourself

Solve x² − 7x + 10 = 0 and verify your roots by back-substitution.

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Factorise: (x − 2)(x − 5) = 0 → x = 2 or x = 5.

Example — simultaneous equations

If you solve 2x + y = 7 and x − y = 1 and get x = 8/3, y = 5/3, substitute back into both equations — not just one. A common error is to substitute back into the equation you used last, which will always be satisfied regardless of whether the answer is correct.

Try it yourself

Solve 2x + 3y = 12 and x − y = 1, then substitute back into both equations to confirm your answer.

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From the second equation: x = 1 + y. Substitute into the first: 2(1+y) + 3y = 12 → 5y = 10 → y = 2, x = 3.

Example — a differential equation

If you claim y = Ae^2x + Be^−x solves y″ − y′ − 2y = 0, differentiate twice and substitute back. The left-hand side should collapse to zero identically, for all x and for any constants A and B.

Try it yourself

Verify by back-substitution that y = Ae^3x + Be^−x is a solution of y″ − 2y′ − 3y = 0.

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Both coefficients vanish identically for all constants A and B.

Differentiating after integration

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If you have computed an indefinite integral, differentiate your answer. If you obtain the original integrand, the integral is correct. This check costs almost nothing — differentiation is almost always easier than integration — and it is completely reliable.

Example — integration by parts

Suppose you compute ∫ x e^x dx and obtain (x − 1)e^x + C. Check by differentiating:

This check catches the most common integration by parts errors — a wrong sign, a missing term — immediately.

Try it yourself

A student claims ∫ x²eˣ dx = (x² − 2x + 2)eˣ + C. Differentiate this answer to verify it is correct.

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Calculator check for definite integrals

Modern graphical calculators and computer algebra systems can evaluate definite integrals numerically. If you have computed a definite integral analytically, ask your calculator for the numerical value and compare. The two numbers should agree to many decimal places.

This check is particularly useful in examinations where a calculator is permitted, and is an immediate catch if you have made an error in the limits or a sign error. It does not confirm that your method is correct, but it confirms that your numerical answer is.

Example

Suppose you have evaluated:

Your calculator should confirm ∫₀¹ x eˣ dx ≈ 1.000. If it returns a different value, you have made an error somewhere — either in the integration or in evaluating the limits.

Try it yourself

Evaluate ∫₀^π/2 cos x dx analytically, then state the numerical value a calculator should confirm.

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Calculator should return 1.000 exactly.

Numerical and order-of-magnitude checks

Before committing to a symbolic answer, ask: is this answer numerically plausible? If the answer to a probability question is greater than 1, or the answer to a distance question is negative, something has gone wrong. This sounds obvious, but surprisingly many exam errors are of exactly this type and go unnoticed because the student did not pause to ask the question.

A slightly stronger version: use your calculator to evaluate your symbolic answer at a specific numerical input, then evaluate the original expression at the same input and compare. If they agree, you have not made a symbolic algebra mistake.

Example — partial fractions

Suppose you decompose:

and find A = 4/3, B = 11/3. Rather than re-doing the algebra, substitute x = 0 into both sides:

One substitution checks the algebra without repeating any of it. Use a value of x that does not coincide with a root of the denominator, and is not zero if you already used x = 0 to find A and B.

Try it yourself

Decompose (2x + 3) / ((x − 1)(x + 2)) into partial fractions, then verify your A and B by substituting x = 0 into both sides.

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Write 2x+3 = A(x+2) + B(x−1). Setting x=1: 5 = 3A → A = 5/3. Setting x=−2: −1 = −3B → B = 1/3.

Graphical and plotting checks

Plotting both sides of an equation, or plotting a function and its claimed derivative or integral, provides a visual check that is independent of all algebraic manipulation. A discrepancy that would be invisible in the algebra often jumps out immediately on a graph.

Useful applications:

Equation solutions: plot both sides separately and confirm the intersections are at your claimed solutions.
Integration: plot the integrand and your claimed antiderivative on the same axes; the antiderivative should have turning points where the integrand crosses zero.
Inequalities: shade the solution set on a graph to confirm it matches the region you computed analytically.

Recovering simpler cases

A-level Further Maths — Mechanics (optional)A-level Further Maths — Further Pure (compulsory)

When you have a general solution that reduces to a simpler known problem by setting some parameter to zero, check that it actually does so. This is one of the most powerful checks available in applied mathematics, because it tests the structure of the solution rather than just its numerical value.

Example — damped harmonic oscillator reduces to SHO

The general solution for a damped oscillator (damping constant k, angular frequency ω₀) is a decaying sinusoid. When you set k = 0, you should recover exactly the solution for simple harmonic motion: x = A cos(ω₀t) + B sin(ω₀t). If your general solution does not reduce correctly, the error is in the general solution.

Example — forced oscillator reduces to free oscillator

A forced oscillator with external driving F cos(Ωt) has both a complementary function and a particular integral. When you set the forcing amplitude F to zero, the particular integral should vanish entirely and you should be left with the unforced solution. If F = 0 leaves a non-zero particular integral, something is wrong.

Example — binomial expansion reduces to a known value

If you have expanded (1 + x)ⁿ as a series and are uncertain whether the coefficients are correct, set x = 1:

The sum of all binomial coefficients must equal 2ⁿ. If your coefficients do not sum to the right value, at least one is wrong.

Try it yourself

Expand (1 + x)⁴. Set x = 1 to check that the coefficients sum to 2⁴ = 16.

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Example — Maclaurin series at x = 0

Any Maclaurin series f(x) = a₀ + a₁x + a₂x² + … should satisfy a₀ = f(0). Setting x = 0 in your series is an immediate check on the constant term.

Try it yourself

A student writes the Maclaurin series for cos x as cos x = 1 − x²/2 + x⁴/24 − … Set x = 0 to check the constant term, and explain why the x⁴ coefficient looks suspicious.

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Setting x = 0: cos(0) = 1, and the series gives 1 ✓ — constant term is correct.

However, the correct Maclaurin series is cos x = 1 − x²/2! + x⁴/4! − … and 4! = 24, so x⁴/24 is actually right. A quick check: 4! = 24 ✓.

Dimensional analysis

A-level Maths — Mechanics (compulsory)A-level Further Maths — Mechanics (optional)

Every quantity in mechanics has dimensions: length [L], mass [M], time [T], and combinations thereof. A correct formula must be dimensionally consistent — the dimensions on both sides of an equation must match, and you cannot add quantities with different dimensions.

Dimensional analysis cannot confirm that a numerical coefficient is correct (it cannot, for example, distinguish 2mv² from ½mv²), but it immediately catches a wide class of structural errors.

Example — kinetic energy

Try it yourself

The kinetic energy can also be written as p²/(2m) where p = mv is linear momentum. Verify this formula is dimensionally consistent with energy.

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Example — period of a pendulum

Suppose you derive T = 2π√(g/l) for the period of a simple pendulum. Check the dimensions:

This gives a frequency, not a period — the formula should be T = 2π√(l/g). The dimensional check catches the inversion immediately.

Try it yourself

A student derives the period of a mass-spring system as T = 2π√(k/m), where k is spring stiffness [M T⁻²] and m is mass [M]. Use dimensional analysis to show this formula is wrong, and state the correct form.

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That is a frequency, not a period. The formula should be T = 2π√(m/k), giving dimensions [T] ✓.

Matrix checks

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Does M · M⁻¹ equal the identity?

If you have computed a matrix inverse, multiply M by M⁻¹ (or M⁻¹ by M — both should work). The result must be the identity matrix I. A single off-diagonal non-zero entry or a wrong diagonal entry immediately identifies an error.

You can also check: det(M) · det(M⁻¹) should equal 1. Since det(M) = 1 and det(M⁻¹) = 1, this is consistent.

Try it yourself

Find the inverse of M = [[1, 2], [3, 7]], then verify M · M⁻¹ = I.

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Are your invariant points actually invariant?

An invariant point p satisfies Mp = p. If you have found an invariant point, apply the matrix to it and check you get the same vector back.

Does your eigenvector actually satisfy the eigenvector equation?

An eigenvector v with eigenvalue λ satisfies Mv = λv. Compute Mv directly and check that it equals λv. This catches both wrong eigenvectors and wrong eigenvalues.

Try it yourself

For M = [[5, 2], [0, 3]], verify that v = (1, 0) is an eigenvector with eigenvalue λ = 5.

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Symmetry and sign checks

Many problems have a built-in symmetry that the answer must respect. If you can identify this symmetry before solving, you have a free check.

Even and odd functions: the integral of an odd function over a symmetric interval is zero. If you obtain a non-zero answer, something is wrong.
Probability: all probabilities must lie in [0, 1] and a complete probability distribution must sum (or integrate) to 1. If yours does not, there is an error somewhere in the distribution.
Sign checks: if a physical quantity must be positive (a mass, a distance, a variance), and your answer is negative, you have made an error.
Parity of a polynomial: if a polynomial equation has only even powers, its roots come in ±pairs. An answer that breaks this symmetry is wrong.

Conservation laws

A-level Maths — Mechanics (compulsory)A-level Further Maths — Mechanics (optional)

In mechanics problems, conservation of energy and conservation of momentum provide independent equations that the solution must satisfy. If you have solved a collision problem using impulse-momentum arguments, check that momentum is conserved. If you have solved a problem using energy methods, check that the energy balance closes.

Example — elastic collision

After computing post-collision velocities u and v for an elastic collision, verify both conservation laws independently:

Try it yourself

Ball A (2 kg, 3 m s⁻¹) collides elastically with stationary ball B (2 kg). After the collision, ball A stops and ball B moves at 3 m s⁻¹. Verify both conservation of momentum and conservation of kinetic energy.

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Solving by two different methods

When time permits, the strongest check of all is to solve the same problem by an entirely different method. If both methods agree, the probability that you have made the same error in both is very small.

Solve a trigonometric equation both graphically and analytically.
Evaluate a determinant by cofactor expansion along two different rows.
Find a volume of revolution both by the disk method and by the shell method.
Solve a system of simultaneous equations by both substitution and elimination.
Compute a derivative using the product rule, then verify it using logarithmic differentiation.

Summary of techniques

Technique	What it checks	Best applied to
Back-substitution	Answer satisfies the original equation	Equations of all types, ODEs
Differentiate after integrating	Antiderivative is correct	Any indefinite integral
Calculator definite integral	Numerical value of a definite integral	Any definite integral (calculator exam)
Evaluative / spot-check	Symbolic algebra at a specific value	Partial fractions, series, identities
Plotting	Visual consistency	Equations, inequalities, integrations
Recovering simpler cases	Structure of a general solution	Mechanics, series, combinatorics
Dimensional analysis	Physical units are consistent	Mechanics, physics-style problems
Matrix × inverse = I	Inverse is correct	Matrix inversion
Eigenvector equation Mv = λv	Eigenvalue and eigenvector are consistent	Eigenvalue problems
Symmetry and sign	Structural plausibility	Any problem with known symmetry
Conservation laws	Energy and momentum balance	Collisions, mechanics
Two independent methods	Complete answer	Any problem where two methods exist